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July 30, 2014

July 30, 2014

Posted by **Chelsea** on Wednesday, September 15, 2010 at 12:23pm.

The top half of the circle x^2 + (y-2)^2 = 4

My final answer is y= 2 +/- sqrt of x-4

However, I circled the positive sign because that's the top half. Is this right? I was not sure whether it was sqrt of x-4 or sqrt of x^2-4

- Calc -
**Reiny**, Wednesday, September 15, 2010 at 1:39pmyou are solving for y, so ...

(y-2)^2 = 4 - x^2

y - 2 = ± √(4-x^2)

since you want the top part

y = + √(4-x^2) + 2

incidentally, √(4-x^2) ≠ 2-x

e.g. √(16 - 4) = √12 ≠ 4-2

- Calc -
**Reiny**, Wednesday, September 15, 2010 at 3:25pmyou are solving for y, so ...

(y-2)^2 = 4 - x^2

y - 2 = ± √(4-x^2)

since you want the top part

y = + √(4-x^2) + 2

incidentally, √(4-x^2) ≠ 2-x

e.g. √(16 - 4) = √12 ≠ 4-2

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